# Evaluating $\lim_{x\rightarrow\infty}(x+1-x)$

$$\lim_{x\rightarrow\infty}(x+1-x)=\lim_{x\rightarrow\infty}1=1$$

can we do like this:

\begin{align}\lim_{x\rightarrow\infty}(x+1-x)&=\lim_{x\rightarrow\infty}x(1+\frac{1}{x})-x\\ &=\lim_{x\rightarrow\infty}x\cdot\lim_{x\rightarrow\infty}(1+\frac{1}{x})-\lim_{x\rightarrow\infty}x\\ &=\lim_{x\rightarrow\infty}(x-x)\\ &=0\\ \end{align}

• Nice, you became familiar with $\infty - \infty$ being undetermined. Read carefully, when $\lim_n(a_n + b_n) = \lim_n a_n + \lim_n b_n$ and when $\lim_n(a_n \cdot b_n) = \lim_n a_n \cdot \lim_n b_n$. Actually, I have a simpler proof for you: $$\lim (x+1 -x) = \lim(x+1) - \lim x = \lim x - \lim x = \lim (x-x) = 0$$
– SBF
Nov 26, 2014 at 16:45
• What proof was that Nov 26, 2014 at 17:08
• That $1=0{{{}}}$
– SBF
Nov 26, 2014 at 17:14
• great you want an upvote? Nov 26, 2014 at 17:24
• I rather meant, that to arrive to a wrong result you can do much less actions. You've used incorrectly limit of a product and of a sum; in my version, using incorrectly limit of a sum already indicates a wrong answer, so that could make you more cautious if you wanted to pay attention.
– SBF
Nov 26, 2014 at 19:07

Note that $\lim_{x\rightarrow\infty} f(x)g(x) = \lim_{x\rightarrow\infty} f(x) \cdot \lim_{x\rightarrow\infty} g(x)$ if all the limits exist and are finite. (But not only if.)
The correct way to compute this limit is $$\lim_{x\rightarrow\infty}[x+1-x]=\lim_{x\rightarrow\infty}1=1$$